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The random variable x has the following continuous probability distrib 08 Jul 2018, 05:27
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The random variable x has the following continuous probability distribution in the range \(0 ≤ x ≤\sqrt{2}\), as shown in the coordinate plane with x on the horizontal axis:



The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).

What is the median of x?


A. \(\frac{\sqrt{2} - 1}{2}\)

B. \(\frac{\sqrt{2}}{4}\)

C. \(\sqrt{2} - 1\)

D. \(\frac{\sqrt{2} + 1}{4}\)

E. \(\frac{\sqrt{2}}{2}\)


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The random variable x has the following continuous probability distrib 08 Jul 2018, 06:03
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Bunuel
The random variable x has the following continuous probability distribution in the range \(0 ≤ x ≤\sqrt{2}\), as shown in the coordinate plane with x on the horizontal axis:



The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).

What is the median of x?


A. \(\frac{\sqrt{2} - 1}{2}\)

B. \(\frac{\sqrt{2}}{4}\)

C. \(\sqrt{2} - 1\)

D. \(\frac{\sqrt{2} + 1}{4}\)

E. \(\frac{\sqrt{2}}{2}\)


Show Spoiler
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An interesting question...

The frequency of x is given by the slanting line and x is between 0 and √2

Now the median will require knowing the centre point of this so formed triangle with x axis and y axis..

It will happen where a line drawn horizontally divides the so formed triangle in two equal areas..

Area of bigger triangle = 1/2 *√2*√2=1
So the are of smaller triangle should be 1/2
And let this happen at distance A from √2. Since the triangles - smaller and bigger - both are similar, the vertical line will also be A.

Area now =1/2 *A*A=1/2....A^2=1...A=1
The x axis is √2 and median lies 1 away from √2, so √2-1

C
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The random variable x has the following continuous probability distrib 08 Jul 2018, 06:15
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Bunuel
The random variable x has the following continuous probability distribution in the range \(0 ≤ x ≤\sqrt{2}\), as shown in the coordinate plane with x on the horizontal axis:

The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).

Show more

Bunuel,

The question statement says \(0 ≤ x ≤\sqrt{2}\) and from the graph, the probability of x being \(\sqrt{2}\) is 0. So I guess the question statement should say \(0 ≤ x <\sqrt{2}\) ?
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The random variable x has the following continuous probability distrib 08 Jul 2018, 06:22
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The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).

Hi chetan2u Sir,
Could you please explain the significance of the highlighted information provided in the question?

And how this info is helpful in solving the question?

Thanking you.
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The random variable x has the following continuous probability distrib 08 Jul 2018, 06:32
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The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).

Hi chetan2u Sir,
Could you please explain the significance of the highlighted information provided in the question?

And how this info is helpful in solving the question?

Thanking you.
Show more

Same wine in different bottle
The point tells us that the left edge of x is 0 and right edge is √2.
Although same info is also stated by 0<=x<=√2, but we should be able to make out the redundant matter and not pay attention on it.
If someone starts spending time between the two statements pointing towards same thing, the Question / test makers aim is achieved.

So read the info as gi en and discard what is not important or what gives you the same wine in a different bottle.
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The random variable x has the following continuous probability distrib 27 Sep 2018, 11:53
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Bunuel
The random variable x has the following continuous probability distribution in the range \(0 ≤ x ≤\sqrt{2}\), as shown in the coordinate plane with x on the horizontal axis:



If Prob(\(x < 0\)) = Prob(\(x > \sqrt{2}\)) = 0, what is the median of x?

A. \(\frac{\sqrt{2} - 1}{2}\) \(\,\,\,\,\,\,\,\) B. \(\frac{\sqrt{2}}{4}\) \(\,\,\,\,\,\,\,\) C. \(\sqrt{2} - 1\) \(\,\,\,\,\,\,\,\) D. \(\frac{\sqrt{2} + 1}{4}\) \(\,\,\,\,\,\,\,\) E. \(\frac{\sqrt{2}}{2}\)
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We are talking about the median of a continuous random variable with a given probability density function... and all this wording is absolutely out-of-GMAT´s-scope!

On the other hand, the same problem could be put into GMAT´s reality (for instance) like that:

Quote:

In the figure above, what is the value of x such that the areas under the line from 0 to x, and from x to \(\sqrt{2}\), coincide?
Show more



\(? = x\,\,\,\,:\,\,\,\,{S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} = {S_{\,{\text{III}}}}\,\,\,\,\,\,\left( * \right)\)

\(\left( * \right)\,\,\,\, \Rightarrow \,\,\,\,{S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} = \frac{{{S_{\,{\text{total}}}}}}{2} = \frac{1}{2}\left( {\frac{{\sqrt 2 \cdot \sqrt 2 }}{2}} \right) = \frac{1}{2}\)

\(\frac{1}{2} = {S_{\,{\text{I}}}} + {S_{\,{\text{II}}}} = \frac{{x \cdot x}}{2} + x\left( {\sqrt 2 - 1} \right) = x\sqrt 2 - \frac{{{x^2}}}{2}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,2x\sqrt 2 - {x^2} = 1\,\)


Now we have at least two possible continuations:


First: test alternatives (easy but not so quick).


Second: solve the second-degree equation (need "insight", but VERY quick), as follows (for instance):

\(2x\sqrt 2 - {x^2} = 1\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{x^2} - 2x\sqrt 2 + 1 = 0\)

\({\text{Sum}}\,\, = \,\,\,2\sqrt 2 \,\,\,\,\,,\,\,\,\,{\text{Product}}\,\, = \,\,\,1\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,{x_1} = \sqrt 2 - 1\,\,,\,\,\,{x_2} = \sqrt 2 + 1\)

\({x_2} > \sqrt 2 \,\,\,\,\,\, \Rightarrow \,\,\,\,\,? = {x_1} = \sqrt 2 - 1\)


This solution follows the notations and rationale taught in the GMATH method.

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fskilnik.
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The random variable x has the following continuous probability distrib 27 Sep 2018, 12:19
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Bunuel,
The question statement says \(0 ≤ x ≤\sqrt{2}\) and from the graph, the probability of x being \(\sqrt{2}\) is 0. So I guess the question statement should say \(0 ≤ x <\sqrt{2}\) ?
Show more

Hi, workout!

I have taken the liberty of answering your question. I hope you both (Bunuel and yourself) don´t mind!

When dealing with continuous random variables, the probability of choosing ANY particular value is ALWAYS zero, therefore it really doesn´t matter if you consider to include, or not, any of the interval extremities...

When dealing with continuous random variables, it is useful to consider intervals, not "points", for example(s):

What is the probability that you will buy a light bulb that will not burn-out in less than 2h?

In this case, you are interested in the interval [0,2] , or ]0,2] , or [0,2[ , or ]0,2[ ...

Curiosity: in this situation, it is commonly considered an exponential distribution (see image below).



(The graphs are related to the density and accumulated functions.)

Regards,
Fabio.
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The random variable x has the following continuous probability distrib 18 Oct 2018, 14:19
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Hi fskilnik

I've a problem with the usage of the term "median" here. For e.g. the median of set A (1,2,100) is 2 although 100 not equal 1. So in our case, the median of x must be the value corresponding to x= root2/2. What's wrong here?
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The random variable x has the following continuous probability distrib 18 Oct 2018, 17:00
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Hi fskilnik

I've a problem with the usage of the term "median" here. For e.g. the median of set A (1,2,100) is 2 although 100 not equal 1. So in our case, the median of x must be the value corresponding to x= root2/2. What's wrong here?
Show more
Hi, HisHo.

Thank you for your interest in the subject, even with my explicit mention that it is out of GMAT´s scope.

(I hope I understood your question. If not, please ask again, no problem at all.)

It is NOT as if x denotes the "position" in a list and y denotes the corresponding value at that position.

In this context, the vertical axis offers probabilities associated with values (intervals of values) in the horizontal ("x") axis.

When we are asked the median value of x, it is really the median value of the possible values of x, taking into account the "frequency" of these values that is given (shown) in the vertical axis. (This sentence will be explained more explicitly below.)

If you understood my last sentence, now it must be clear the answer is not:

(1) y=f(a) , where a is (square root of 2)/2
(2) y=f(b) , for any other b

In both cases because y does not have this "value" property associated with the position that would be given in the x axis.
(This is not what is going on, so to speak!)

(3) (square root of 2)/2 , because the "distribution of frequencies" is "irregular", I mean, the smaller (positive) values of x are more frequent...
Consequence: the value x=m (the median) that divides the "list" of x´s (the whole interval from 0 to the square root of 2) such that,
the probability of randomly choosing a number in the interval [0,m] is 50% , is less the (square root of 2)/2 ... got it?

I hope you understood (and liked) my INFORMAL comments.

Regards and success in your studies,
Fabio.
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The random variable x has the following continuous probability distrib 16 Oct 2019, 18:49
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Can someone provide please an additional explanation?

Kind regards!
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The random variable x has the following continuous probability distrib 07 Apr 2020, 22:56
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Can I assume that the area under the curve of x+y=sqrt(2) is 1, so the question needs us to find the interval from [0,a] such that the probability under the curve is 50%, and we could use the fundamental theorem of calculus (i.e. integral) to calculate the endpoint?
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The random variable x has the following continuous probability distrib 11 Jun 2020, 21:51
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chetan2u
Bunuel
The random variable x has the following continuous probability distribution in the range \(0 ≤ x ≤\sqrt{2}\), as shown in the coordinate plane with x on the horizontal axis:


The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).

What is the median of x?


A. \(\frac{\sqrt{2} - 1}{2}\)

B. \(\frac{\sqrt{2}}{4}\)

C. \(\sqrt{2} - 1\)

D. \(\frac{\sqrt{2} + 1}{4}\)

E. \(\frac{\sqrt{2}}{2}\)


Show Spoiler
Attachment:
BK8gXRh.jpg

An interesting question...

The frequency of x is given by the slanting line and x is between 0 and √2

Now the median will require knowing the centre point of this so formed triangle with x axis and y axis..

It will happen where a line drawn horizontally divides the so formed triangle in two equal areas..

Area of bigger triangle = 1/2 *√2*√2=1
So the are of smaller triangle should be 1/2
And let this happen at distance A from √2. Since the triangles - smaller and bigger - both are similar, the vertical line will also be A.

Area now =1/2 *A*A=1/2....A^2=1...A=1
The x axis is √2 and median lies 1 away from √2, so √2-1

C
Show more

Thank you, this is a great response. I was trying to understanding how come the smaller triangle is isosceles. Just to add further details to your explanation for those who still wonder, the smaller triangle is SIMILAR to the large triangle since they're having the same angles. And because the large triangle is isosceles (square root of 2) so is the smaller one.
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The random variable x has the following continuous probability distrib 14 Jul 2020, 21:28
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chetan2u
Bunuel
The random variable x has the following continuous probability distribution in the range \(0 ≤ x ≤\sqrt{2}\), as shown in the coordinate plane with x on the horizontal axis:



The probability that x < 0 = the probability that \(x > \sqrt{2} = 0\).

What is the median of x?


A. \(\frac{\sqrt{2} - 1}{2}\)

B. \(\frac{\sqrt{2}}{4}\)

C. \(\sqrt{2} - 1\)

D. \(\frac{\sqrt{2} + 1}{4}\)

E. \(\frac{\sqrt{2}}{2}\)


Show Spoiler
Attachment:
BK8gXRh.jpg

An interesting question...

The frequency of x is given by the slanting line and x is between 0 and √2

Now the median will require knowing the centre point of this so formed triangle with x axis and y axis..

It will happen where a line drawn horizontally divides the so formed triangle in two equal areas..

Area of bigger triangle = 1/2 *√2*√2=1
So the are of smaller triangle should be 1/2
And let this happen at distance A from √2. Since the triangles - smaller and bigger - both are similar, the vertical line will also be A.

Area now =1/2 *A*A=1/2....A^2=1...A=1
The x axis is √2 and median lies 1 away from √2, so √2-1

C
Show more

can you elaborate on "Now the median will require knowing the centre point of this so formed triangle with x axis and y axis..

It will happen where a line drawn horizontally divides the so formed triangle in two equal areas.."

What is the logic for this.
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The random variable x has the following continuous probability distrib 31 May 2025, 04:50
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